《ACM程序设计》书中题目M

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本文介绍了一个用于解决特定数学问题的程序：在一个由2到15个唯一正整数组成的列表中找出多少个数是另一个数的两倍。程序通过读取输入文件中的多个列表，使用排序和双循环来确定每组数据的答案。

Description
As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and asked to determine how many items in each list are twice some other item in the same list. You will need a program to help you with the grading. This program should be able to scan the lists and output the correct answer for each one. For example, given the list
1 4 3 2 9 7 18 22

your program should answer 3, as 2 is twice 1, 4 is twice 2, and 18 is twice 9.

Input

The input file will consist of one or more lists of numbers. There will be one list of numbers per line. Each list will contain from 2 to 15 unique positive integers. No integer will be larger than 99. Each line will be terminated with the integer 0, which is not considered part of the list. A line with the single number -1 will mark the end of the file. The example input below shows 3 separate lists. Some lists may not contain any doubles.

Output

The output will consist of one line per input list, containing a count of the items that are double some other item.

Sample Input

1 4 3 2 9 7 18 22 0
2 4 8 10 0
7 5 11 13 1 3 0
-1

Sample Output

3
2
0

这题找成2倍关系数的个数，用vector存比较合适，记得每次最后清空向量

代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
    vector<int>v;
    int cmp;
    int n;
    while(1)
    {

        for(int i=0;i<20;i++)
        {           

            scanf("%d",&cmp);
            if(cmp==0)
                break;
            v.push_back(cmp);

        }
        if(cmp==-1)
            break;
        sort(v.begin(),v.end());
        int cnt=0;
        for(int j=0;j<v.size();j++)
        {
            for(int k=j+1;k<v.size();k++)
            {
                if(v[j]*2==v[k])
                {
                    cnt++;
                }

            }
        }
        printf("%d\n",cnt);
        v.clear();
    }
    return 0;   
}